Problem: Let $h(x)=\dfrac{\sqrt{x+12}-3}{x+3}$ when $x\neq -3$. $h$ is continuous for all $x>-12$. Find $h(-3)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{6}$ (Choice B) B $-\dfrac{1}{9}$ (Choice C) C $\dfrac{1}{9}$ (Choice D) D $-\dfrac{1}{6}$
Answer: $\dfrac{\sqrt{x+12}-3}{x+3}$ is continuous for all $x>-12$ other than $x=-3$, which means $h$ is continuous for all $x>-12$ other than $x=-3$. In order for $h$ to also be continuous at $x=-3$, the following equality must hold: $\lim_{x\to -3}h(x)=h(-3)$ We will obtain the above equality by letting $h(-3)=\lim_{x\to -3}h(x)$. So let's find $\lim_{x\to -3}h(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -3}h(x) \\\\ &=\lim_{x\to -3}\dfrac{\sqrt{x+12}-3}{x+3} \gray{\text{This is the rule for }x\neq -3} \\\\ &=\lim_{x\to -3}\dfrac{\sqrt{x+12}-3}{x+3}\cdot\dfrac{\sqrt{x+12}+3}{\sqrt{x+12}+3} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to -3}\dfrac{x+12-3^2}{(x+3)(\sqrt{x+12}+3)} \gray{\text{Simplify}} \\\\ &=\lim_{x\to -3}\dfrac{\cancel{x+3}}{\cancel{(x+3)}(\sqrt{x+12}+3)} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to -3}\dfrac{1}{(\sqrt{x+12}+3)} \\\\ &\text{(This is allowed because }x\neq -3) \\\\ &=\dfrac{1}{\sqrt{-3+12}+3} \gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{6} \end{aligned}$ We obtained that if we set $h(-3)=\dfrac{1}{6}$, then $\lim_{x\to -3}h(x)=h(-3)$, which makes $h$ continuous at $x=-3$. Since we already saw that $h$ is continuous for any other $x>-12$, we can determine that it's continuous for all $x>-12$. In conclusion, $k=\dfrac{1}{6}$.